(7t-2)(3t+1)=3(13t)

Solution for (7t-2)(3t+1)=3(13t) equation:

(7t-2)(3t+1)=3(13t)

We move all terms to the left:

(7t-2)(3t+1)-(3(13t))=0

determiningTheFunctionDomain
(7t-2)(3t+1)-313t=0

We add all the numbers together, and all the variables

-313t+(7t-2)(3t+1)=0

We multiply parentheses ..

(+21t^2+7t-6t-2)-313t=0

We get rid of parentheses

21t^2+7t-6t-313t-2=0

We add all the numbers together, and all the variables

21t^2-312t-2=0

a = 21; b = -312; c = -2;
Δ = b2-4ac
Δ = -3122-4·21·(-2)
Δ = 97512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{97512}=\sqrt{4*24378}=\sqrt{4}*\sqrt{24378}=2\sqrt{24378}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-312)-2\sqrt{24378}}{2*21}=\frac{312-2\sqrt{24378}}{42}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-312)+2\sqrt{24378}}{2*21}=\frac{312+2\sqrt{24378}}{42}
$